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Is k^2 +k -2 > 0 ?

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Is k^2 +k -2 > 0 ? [#permalink] New post 12 Feb 2013, 15:52
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Is k^2 + k - 2 > 0 ?

(1) k < 1
(2) k < -2
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Feb 2013, 15:57, edited 1 time in total.
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Re: Is k^2 +k -2 > 0 ? [#permalink] New post 12 Feb 2013, 16:11
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Is k^2 + k - 2 > 0 ?

First let's see for which ranges of k the given inequality holds true: k^2 + k - 2 > 0 --> (k+2)(k-1)>0 --> roots are -2 and 1 --> ">" sign indicates that the solution lies to the left of a smaller root and to the right of a larger root (check here for this technique: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476). Thus the given inequality holds true for: k<-2 and k>1.

So, the question asks whether k<-2 or k>1.

(1) k < 1. Not sufficient.
(2) k < -2. Sufficient.

Answer: B.
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Re: Is k^2 +k -2 > 0 ? [#permalink] New post 12 Feb 2013, 16:11
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rephrasing the statement we have

(x + 2) (x - 1) > 0 ====> x > -2 AND x > 1

on a number line we have

-------------------0 ----- 1 +++++++

----- - 2 +++++0++++++++++++

So from 1) we have x > 1 so 2 intervals : one positive and one negative. two values. not suff

from2) x < - 2 the only values are positive. so is suff.

B is the answer

Bunuel ?? correct my approach, pretty straightforward :)
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Re: Is k^2 +k -2 > 0 ?   [#permalink] 12 Feb 2013, 16:11
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